Question

An ideal solenoid having 40 turns ${\text{cm}}^{-1}$ has an aluminum core, and carries a current of $2.0\text{A}$. Calculate the magnetic field $B$ at the centre. The susceptibility $\chi$ of aluminium $=2.3\times {10}^{-5}$.

• A. $3.2\pi \times {10}^{-4}\text{T}$
• B. $1.6\pi \times {10}^{-4}\text{T}$
• C. $0.8\pi \times {10}^{-4}\text{T}$
• D. $\pi \times {10}^{-4}\text{T}$

Answer 1

The correct option is A $3.2\pi \times {10}^{-4}\text{T}$

$n=40$ turns ${\text{cm}}^{-1}=4000\text{turns}{\text{m}}^{-1}$

$i=2A$, $\chi =2.3\times {10}^{-5}$

Magnetization is given by,

$I=\chi H=\chi ni=(2.3\times {10}^{-5}\times 4000\times 2)$

$I=0.18{\text{Am}}^{-1}$

Now, $B={\mu }_0(H+I)$

$B=4\pi \times {10}^{-7}(8000+0.18)$

$B=3.2\pi \times {10}^{-4}\text{T}$

Hence, option $\left(A\right)$ is the correct answer.